TITLE: prove $s(x+y)=s(x)s(y)$
QUESTION [5 upvotes]: I am asked to prove the following:

Let $s(x):=\sum_{n=0}^{\infty}\binom{x}{n}$. Then $s(x+y)=s(x)s(y)$.

I don't know how to start. I am thinking about $\exp(x)$ function with $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$. Am I okay with this start? I would then begin like this: 
$$\sum_{n=0}^{n}\dfrac{x^n}{n!} \cdot \dfrac{y^n}{n!} = .. help = \frac{1}{n!}(x+y)^n$$
I am not sure how to continue or whether I am okay. 
Thanks for help.

REPLY [4 votes]: We have the generalized binomial theorem
$$ (1+z)^x=  \sum_{n=0}^\infty \binom{x}{n}z^n $$
where $\binom{x}{n}$ denotes the generalized binomial coefficient
$$ \binom{x}{n} = \frac{x(x-1)(x-2)\cdot\ldots\cdot(x-n+1)}{n}$$
Provided $x > -1$, this series converges with radius of convergence $1$, and specifically for $z=1$; convergence at $z=1$ is absolute if and only if $x > 0$. Thus, assuming $x > 0$, we have $s(x)=(1+1)^x=2^x$, and now the problem is trivial:  $$s(x)s(y)=2^x2^y=2^{x+y}=s(x+y)$$ for $x,y>0$.